# Computing $\sqrt[4]{28+16 \sqrt 3}$

I want to compute following radical

$$\sqrt[4]{28+16 \sqrt 3}$$

For that, I first tried to rewrite this in terms of exponential.

$$(28+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$

We know that $$28 = 2 \cdot 7^{\frac{1}{2}}$$

$$(2 \cdot 7^{\frac{1}{2}}+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$

However, I'm stuck at this step. Could you assist me?

Regards

• Is there an exponential approach to this question?
– Melz
Dec 22, 2018 at 10:08
• The statment $28 = 2 \cdot 7^{\frac{1}{2}}$ is false. It should be $28=2^2\cdot 7$ or $28^{\frac12} = 2\cdot 7^{\frac{1}{2}}$. Dec 22, 2018 at 10:12

$$\sqrt[4] {28+16\sqrt 3}=\sqrt[4] {(\sqrt {12})^2 +(\sqrt {16})^2 +2\sqrt {16\cdot 12}}=\sqrt {4+2\sqrt 3}=\sqrt {(\sqrt 3)^2 +(1)^2 +2\sqrt {3\cdot 1}}=\sqrt 3 +1$$

• Is it a rule or something?
– Melz
Dec 22, 2018 at 10:05
• @Enzo Since you are taking the fourth root, you would like to express $28+16\sqrt3$ as the fourth power of some number, or the square of the square of some number. In other words, you want to find $a,b$ such that $(a+b)^2=a^2+b^2+2ab=28+16\sqrt3$, and then $c,d$ such that $(c+d)^2=a+b$ Dec 22, 2018 at 10:11

Hint.

• $$28+16\sqrt3=12+16+2\cdot2\sqrt3\cdot4=(2\sqrt3)^2+4^2+2\cdot2\sqrt3\cdot4=(4+2\sqrt3)^2$$

• $$4+2\sqrt3=3+1+2\cdot1\cdot\sqrt3=(\sqrt3)^2+1^2+2\cdot1\cdot\sqrt3=(\sqrt3+1)^2$$

Hint:

Try to write $$28+16 \sqrt 3=(a+b\sqrt 3)^4$$ for suitable $$a$$ and $$b$$.

You obtain, reordering the terms \begin{align} (a+b\sqrt 3)^4&=(a^4+18a^2b^2+9b^4)+4ab(a^2+3b^2)\sqrt 3.\\ \end{align} Can you choose $$a$$ and $$b$$ so that $$a^4+18a^2b^2+9b^4=28,\quad ab(a^2+3b^2)=4?$$

To find the fourth root, we need to use Bill Dubuque's denesting algorithm for square roots twice.

First time round, we can find that $$\sqrt{\text{norm}} = \sqrt{28^2 - 3 \cdot 16^2} = \sqrt{16} = 4$$. Subtracting out the norm gives $$24 + 16 \sqrt{3}$$, and when you divide this by $$\sqrt{\text{trace}} = \sqrt{2 \cdot 24} = 4 \sqrt{3}$$, we get $$\frac{8 \sqrt{3}\sqrt 3}{4 \sqrt 3} + \frac{16 \sqrt3}{4 \sqrt3} = 2\sqrt{3}+4 = 4 + 2\sqrt{3}$$.

Second time round, we can find that $$\sqrt{\text{norm}} = \sqrt{4^2 - 3 \cdot 2^2 -8} = \sqrt{4} = 2$$. Subtracting out the norm gives $$2 + 2\sqrt{3}$$, and when you divide this by $$\sqrt{\text{trace}} = \sqrt{4} = 2$$, you get $$1 + \sqrt{3}$$.

Note

$$\sqrt[4]{28+16 \sqrt 3}=\sqrt[4]{4(7+2 \sqrt {3\cdot4})} = \sqrt[4]{4(\sqrt4+\sqrt {3})^2}\\ = \sqrt{4+2\sqrt{1\cdot3}}=\sqrt{(1+\sqrt3)^2}=1+\sqrt3$$

In general, there's not much you can do with an expression like that. But if you know that the answer is in the form $$a+b\sqrt 3$$ for integers $$a$$ and $$b$$, then you can solve it like this:

Step 1

($$a+b\sqrt 3)^4=28+16\sqrt 3$$. So suppose ($$a+b\sqrt 3)^2=c+d\sqrt 3$$. Then $$(c+d\sqrt 3)^2=28+16\sqrt 3$$, i.e.

$$(c^2+3d^2)+2cd\sqrt 3=28+16\sqrt 3$$

$$c^2+3d^2$$ and $$2cd$$ are integers, so this can only be true if $$c^2+3d^2=28$$ and $$2cd=16$$.
Then $$d=8/c$$, so $$c^2+\dfrac{192}{c^2}=28$$. Multiplying by $$c^2$$ and rearranging, we get a quadratic in $$c^2$$:

$$c^4-28c^2+192=0$$

This factors as $$(c^2-12)(c^2-16)=0$$. Since $$c$$ is an integer, we must have $$c=\pm4$$, and so $$d=\pm2$$. So we have two possible solutions: $$\pm(4+2\sqrt 3)$$. (You should try squaring this to check that you get the expected result $$28+16\sqrt 3$$). Pick the positive solution $$4+2\sqrt 3$$ (if you choose the negative solution, you will find that Step 2 fails).

Step 2

Now you need to solve $$(a+b\sqrt 3)^2=4+2\sqrt 3$$. You should be able to do this yourself now, using exactly the same method as Step 1.

Alternatively, you can use Michael Rozenberg's identities, where this identity applies to the question:

$$\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$

We have $$a = 28$$, $$b = 16^2 \cdot 3 = 768$$. Applying this identity gives:

$$\sqrt{28+\sqrt{768}}=\sqrt{\frac{28+\sqrt{16}}{2}}+\sqrt{\frac{28-\sqrt{16}}{2}}=\sqrt{16}+\sqrt{12}=4+2\sqrt{3}.$$

Applying the identity once more gives, where $$a=4, b=2^2 \cdot 3 = 12$$ gives: $$\sqrt{4+\sqrt{12}}=\sqrt{\frac{4+\sqrt{4^2-12}}{2}}+\sqrt{\frac{4-\sqrt{4^2-12}}{2}}=\sqrt{\frac{4+2}{2}}+\sqrt{\frac{4-2}{2}}=\sqrt{3}+1.$$